The correct option is
D (3,4,5)The equation of line through B(1,4,6) and C(5,4,4) is
x−15−1=y−44−4=z−64−6
or x−14=y−40=z−6−2=k
or x−1=4k,y−4=0,z−6=−2k
or x=4k+1,y=4,z=−2k+6
Any point D on it is (4k+1,4,−2k+6)
Let D be the foot of the perpendicular from A on BC
Direction ratios of AD are 4k+1−1,4−2,−2k+6−1
Direction ratios of AD are 4k,2,−2k+5
Direction ratios of BC are 4,0,−2
Since AD is perpendicular to BC
∴4k×4+2×0+(−2k+5)(−2)=0
∴16k+0+4k−10=0
∴20k=10 or k=12
∴D is (2+1,4,−1+6) or (3,4,5)